3.99.51 \(\int \frac {500 x^2+e^{5/x} (150000+33000 x+1815 x^2)}{30000 x^2+6600 x^3+363 x^4} \, dx\)

Optimal. Leaf size=25 \[ 8-e^{5/x}+\frac {x}{3 \left (20+\frac {11 x}{5}\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 21, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {1594, 27, 12, 6742, 2209} \begin {gather*} -e^{5/x}-\frac {500}{33 (11 x+100)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(500*x^2 + E^(5/x)*(150000 + 33000*x + 1815*x^2))/(30000*x^2 + 6600*x^3 + 363*x^4),x]

[Out]

-E^(5/x) - 500/(33*(100 + 11*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{x^2 \left (30000+6600 x+363 x^2\right )} \, dx\\ &=\int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{3 x^2 (100+11 x)^2} \, dx\\ &=\frac {1}{3} \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{x^2 (100+11 x)^2} \, dx\\ &=\frac {1}{3} \int \left (\frac {15 e^{5/x}}{x^2}+\frac {500}{(100+11 x)^2}\right ) \, dx\\ &=-\frac {500}{33 (100+11 x)}+5 \int \frac {e^{5/x}}{x^2} \, dx\\ &=-e^{5/x}-\frac {500}{33 (100+11 x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 27, normalized size = 1.08 \begin {gather*} \frac {5}{3} \left (-\frac {3 e^{5/x}}{5}-\frac {100}{11 (100+11 x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(500*x^2 + E^(5/x)*(150000 + 33000*x + 1815*x^2))/(30000*x^2 + 6600*x^3 + 363*x^4),x]

[Out]

(5*((-3*E^(5/x))/5 - 100/(11*(100 + 11*x))))/3

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 24, normalized size = 0.96 \begin {gather*} -\frac {33 \, {\left (11 \, x + 100\right )} e^{\frac {5}{x}} + 500}{33 \, {\left (11 \, x + 100\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1815*x^2+33000*x+150000)*exp(5/x)+500*x^2)/(363*x^4+6600*x^3+30000*x^2),x, algorithm="fricas")

[Out]

-1/33*(33*(11*x + 100)*e^(5/x) + 500)/(11*x + 100)

________________________________________________________________________________________

giac [A]  time = 0.14, size = 32, normalized size = 1.28 \begin {gather*} -\frac {\frac {300 \, e^{\frac {5}{x}}}{x} + 33 \, e^{\frac {5}{x}} - 5}{3 \, {\left (\frac {100}{x} + 11\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1815*x^2+33000*x+150000)*exp(5/x)+500*x^2)/(363*x^4+6600*x^3+30000*x^2),x, algorithm="giac")

[Out]

-1/3*(300*e^(5/x)/x + 33*e^(5/x) - 5)/(100/x + 11)

________________________________________________________________________________________

maple [A]  time = 0.08, size = 17, normalized size = 0.68

method	result	size
risch	\(-\frac {500}{363 \left (x +\frac {100}{11}\right )}-{\mathrm e}^{\frac {5}{x}}\)	\(17\)
derivativedivides	\(\frac {5}{3 \left (\frac {100}{x}+11\right )}-{\mathrm e}^{\frac {5}{x}}\)	\(21\)
default	\(\frac {5}{3 \left (\frac {100}{x}+11\right )}-{\mathrm e}^{\frac {5}{x}}\)	\(21\)
norman	\(\frac {-\frac {500 x}{33}-100 x \,{\mathrm e}^{\frac {5}{x}}-11 x^{2} {\mathrm e}^{\frac {5}{x}}}{x \left (11 x +100\right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1815*x^2+33000*x+150000)*exp(5/x)+500*x^2)/(363*x^4+6600*x^3+30000*x^2),x,method=_RETURNVERBOSE)

[Out]

-500/363/(x+100/11)-exp(5/x)

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 18, normalized size = 0.72 \begin {gather*} -\frac {500}{33 \, {\left (11 \, x + 100\right )}} - e^{\frac {5}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1815*x^2+33000*x+150000)*exp(5/x)+500*x^2)/(363*x^4+6600*x^3+30000*x^2),x, algorithm="maxima")

[Out]

-500/33/(11*x + 100) - e^(5/x)

________________________________________________________________________________________

mupad [B]  time = 5.60, size = 18, normalized size = 0.72 \begin {gather*} -{\mathrm {e}}^{5/x}-\frac {500}{33\,\left (11\,x+100\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5/x)*(33000*x + 1815*x^2 + 150000) + 500*x^2)/(30000*x^2 + 6600*x^3 + 363*x^4),x)

[Out]

- exp(5/x) - 500/(33*(11*x + 100))

________________________________________________________________________________________

sympy [A]  time = 0.17, size = 12, normalized size = 0.48 \begin {gather*} - e^{\frac {5}{x}} - \frac {500}{363 x + 3300} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1815*x**2+33000*x+150000)*exp(5/x)+500*x**2)/(363*x**4+6600*x**3+30000*x**2),x)

[Out]

-exp(5/x) - 500/(363*x + 3300)

________________________________________________________________________________________